import smtplib
from email.mime.multipart import MIMEMultipart
from email.mime.text import MIMEText

def sendEmail(to,htmlText,
              subject='No Subject',
              me ='CIgrino@consumerintel.com',
              replyTo='iliask@consumerintel.com',
              cc='iliask@consumerintel.com',
              server='mail.consumerintel.co.uk'):
    # me == my email address
    # you == recipient's email address
    
    # Create message container - the correct MIME type is multipart/alternative.
    msg = MIMEMultipart('alternative')
    msg['Subject'] = subject
    msg['From'] = me
    msg['To'] = to
    msg['Reply-to']=replyTo
    msg['Cc'] = cc 
    
    
    # Create the body of the message (a plain-text and an HTML version).
#    text = "Hi!\nHow are you?\nHere is the link you wanted:\nhttp://www.python.org"
#    html = """\
#    <html>
#      <head></head>
#      <body>
#        <p>Hi!<br>
#           How are you?<br>
#           Here is the <a href="http://www.python.org">link</a> you wanted.
#        </p>
#      </body>
#    </html>
#    """
    html=htmlText
    # Record the MIME types of both parts - text/plain and text/html.
    #part1 = MIMEText(text, 'plain')
    part2 = MIMEText(html, 'html')
    
    # Attach parts into message container.
    # According to RFC 2046, the last part of a multipart message, in this case
    # the HTML message, is best and preferred.
    #msg.attach(part1)
    msg.attach(part2)
    
    # Send the message via local SMTP server.
    s = smtplib.SMTP(server)
    # sendmail function takes 3 arguments: sender's address, recipient's address
    # and message to send - here it is sent as one string.
    s.sendmail(me, to, msg.as_string())
    if cc:
        s.sendmail(me, cc, msg.as_string())
    s.quit()
if __name__=='__main__':
    sendEmail(to='Kirsty.Anthony-johns@consumerintel.com',
              htmlText='Test',
              subject='Test'
              #,cc='iliask@consumerintel.com'
              )